The energy density in a parallel plate capacitor is given as 2.1*10^-9 J/m^3.The value of electric field in the region between the plate is ?

Question

Answer

Calculating Electric Field in a Parallel Plate Capacitor

Given Information

Energy density in the capacitor: 2.1 x 10^-9 J/m^3

Formula for energy density in a capacitor: U = (1/2)εE^2

Where U is the energy density, ε is the permittivity of the material between the plates, and E is the electric field strength

Permittivity of free space, ε0 = 8.85 x 10^-12 C^2/Nm^2

Solving for Electric Field Strength

We can rearrange the formula for energy density to solve for the electric field strength:

E = sqrt(2U/ε)

Plugging in the given values:

E = sqrt(2 x 2.1 x 10^-9 J/m^3 / 8.85 x 10^-12 C^2/Nm^2)

E = 191,770 N/C

Explanation

In a parallel plate capacitor, the electric field strength is constant between the plates and is given by E = V/d, where V is the potential difference between the plates and d is the distance between them. However, in this problem, we are given the energy density instead of the potential difference. We can use the formula for energy density, which relates the energy stored in the capacitor to the electric field strength, to solve for E. The permittivity of the material between the plates is also needed, which is typically given in the problem or can be looked up in a table.

Once the electric field strength is calculated, it can be used to find the potential difference between the plates and the charge on each plate using the equations V = Ed and Q = CV, where C is the capacitance of the capacitor. These calculations are not necessary for this problem, but may be useful in other capacitor problems.

Answer

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